$\overline{AC} = 9$ $\overline{BC} = {?}$ $A$ $C$ $B$ $9$ $?$ $ \sin( \angle ABC ) = \frac{3\sqrt{10} }{10}, \cos( \angle ABC ) = \frac{ \sqrt{10}}{10}, \tan( \angle ABC ) = 3$
Solution: $\overline{AC}$ is the opposite to $\angle ABC$ $\overline{BC}$ is adjacent to $\angle ABC$ SOH CAH TOA We know the opposite side and need to solve for the adjacent side so we can use the tan function (TOA) $ \tan( \angle ABC ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\overline{AC}}{\overline{BC}}= \frac{9}{\overline{BC}} $ $ \overline{BC}=\frac{9}{\tan( \angle ABC )} = \frac{9}{3} = 3$